Unitary $T : V \to V$ preserves $W \le V$ and $W^\perp$.

Question

Let $V$ be an inner product space over $\mathbb{F}$, where $\mathbb{F}$ is $\mathbb{R}$ or $\mathbb{C}$, and $T : V \to V$ a unitary linear operator, i.e. $T \circ T^* = \mathrm{id} = T^* \circ T$, where $T^*$ is the adjoint of $T$. Suppose that $W \le V$ is a $T$-invariant subspace, that is $TW \subseteq W$. Show that the orthogonal complement $W^\perp$ is also $T$-invariant.

I have a solution for when $V$ is finite-dimensional, which I have posted as an answer below. However, I would be interested in knowing if the result also holds in infinite dimensions, and seeing a proof thereof. Alternative proofs for finite dimensions are also welcome, though I shan't accept them as answers.

Answer 1

This is not true in infinite dimension. Take for example $l^2(\mathbb{Z})$ and $T$ the left shift operator $T(f)(n)=f(n+1)$ and $V=\{f:f(n)=0\text{ for }n>0\}$. Then $T$ is unitary, $V$ is $T$-invariant, but $V^\perp$ is not $T$-invariant.

Answer 2

Suppose further that $V$ is finite-dimensional and, without loss of generality, that $W \ne V$. Let $w \in W^\perp$ be non-zero and consider the sequence $(T^k w)_{k \in \mathbb{N}}$. By finite-dimensionality, there exists a non-trivial linear dependence $$\sum_{k=0}^n \lambda_k T^k w = 0,$$ for some $n \in \mathbb{Z}$ and $\lambda_k \in \mathbb{F}$. Since $w \ne 0$, we may choose $n > 0$ so that $\lambda_n \ne 0$. Then, by applying $(T^*)^{n-1}$ and rearranging the above, we find that $$\lambda_n Tw = -\sum_{k=0}^{n-1} \lambda_k (T^*)^{n-1-k} w.$$ But $W^\perp$ is $T^*$-invariant, so $(T^*)^k w \in W^\perp$ for all $k \ge 0$. Since $\lambda_n \ne 0$, we conclude that $Tw \in W^\perp$. Hence $W^\perp$ is $T$-invariant.